a2 appendix a vectors - cengage · pdf filea2 appendix a vectors the term vector is used by...

A2 APPENDIX A VECTORS

The term vector is used by scientists to indicate a quantity (such as displacement or veloc-ity or force) that has both magnitude and direction. A vector is often represented by an arrowor a directed line segment. The length of the arrow represents the magnitude of the vectorand the arrow points in the direction of the vector. We denote a vector by printing a letterin boldface or by putting an arrow above the letter

For instance, suppose a particle moves along a line segment from point to point .The corresponding displacement vector , shown in Figure 1, has initial point (the tail)and terminal point (the tip) and we indicate this by writing AB

l. Notice that the vec-

tor CDl

has the same length and the same direction as even though it is in a differentposition. We say that and are equivalent (or equal) and we write . The. zero vec-tor, denoted by 0, has length . It is the only vector with no specific direction.

Combining VectorsSuppose a particle moves from , so its displacement vector is AB

l. Then the particle

changes direction and moves from , with displacement vector BCl

as in Figure 2. Thecombined effect of these displacements is that the particle has moved from . Theresulting displacement vector AC

lis called the sum of AB

land BC

land we write

ACl

ABl

BCl

In general, if we start with vectors and , we first move so that its tail coincides withthe tip of and define the sum of and as follows.

�vl�.�v�BA

Avv �B

vu �u � vvu

0

A to BB to C

A to C

��

vvuvuu

A.1 Vectors in Two Dimensions

FIGURE 1Equivalent vectors

A

B

v

C

D

u

FIGURE 2

C

B

A

48238_ch10_ptg01_hr_670-679_48328_ch10_ptg01_hr_670-679 1/24/11 5:20 PM Page 678

Definition of Vector Addition If and are vectors positioned so the initial point ofis at the terminal point of , then the sum is the vector from the initial

point of to the terminal point of .

The definition of vector addition is illustrated in Figure 3. You can see why this defi ni-tion is sometimes called the Triangle Law.

In Figure 4 we start with the same vectors and as in Figure 3 and draw another copy of with the same initial point as . Completing the parallelogram, we see that

. This also gives another way to construct the sum: If we place and sothey start at the same point, then lies along the diagonal of the parallelogram withand as sides. (This is called the Parallelogram Law.)

u vv

vuu � vu

FIGURE 3 The Triangle Law

vu+v

u

FIGURE 4 The Parallelogram Law

v v+u

u

uv

u+v

u vv u

u � v � v � u u vu � v u

v

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SECTION A.1 VECTORS IN TWO DIMENSIONS A3

Draw the sum of the vectors shown in Figure 5.

SOLUTION First we translate and place its tail at the tip of , being careful to draw acopy of that has the same length and direction. Then we draw the vector [seeFigure 6(a)] starting at the initial point of and ending at the terminal point of the copyof .

Alternatively, we could place so it starts where starts and construct by theParallelogram Law as in Figure 6(b).

It is possible to multiply a vector by a real number . (In this context we call the real num-ber a scalar to distinguish it from a vector.) For instance, we want to be the same vec-tor as , which has the same direction as but is twice as long. In general, we multiplya vector by a scalar as follows.

Definition of Scalar Multiplication If is a scalar and is a vector, then the scalarmultiple is the vector whose length is times the length of and whosedirection is the same as if and is opposite to if . If or ,then .

a and b

b ab a � b

ab

b a a � b

FIGURE 6

a

b

a+b

(a)

a

a+bb

(b)

cc 2v

EXAMPLE 1

v � v v

c vcv c v

v c � 0 v c � 0 c � 0 v � 0cv � 0

FIGURE 5

a b


This definition is illustrated in Figure 7. We see that real numbers work like scaling fac-tors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel ifthey are scalar multiples of one another. In particular, the vector has the samelength as but points in the opposite direction. We call it the negative of .

By the difference of two vectors we mean

So we can construct by first drawing the negative of , , and then adding it toby the Parallelogram Law as in Figure 8(a). Alternatively, since the vec-tor , when added to , gives . So we could construct as in Fig ure 8(b) bymeans of the Triangle Law.

�v � ��1�vvv

u � v

u � v � u � ��v�

u�vvu � vv � �u � v� � u,

u � vuvu � v

FIGURE 8Drawing u-v (a)

uv

u-v

_v

(b)

v

u-v

u

_1.5v

v 2v

_v

v12

FIGURE 7Scalar multiples of v

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A4 APPENDIX A VECTORS

If are the vectors shown in Figure 9, draw .

SOLUTION We first draw the vector pointing in the direction opposite to and twiceas long. We place it with its tail at the tip of and then use the Triangle Law to draw

as in Figure 10.

Components

For some purposes it’s best to introduce a coordinate system and treat vectors algebra-ically. If we place the initial point of a vector at the origin of a rectangular two-dimen-sional coordinate system, then the terminal point of has coordinates of the form(See Figure 11.) These coordinates are called the components of and we write

We use the notation for the ordered pair that refers to a vector so as not to confuseit with the ordered pair that refers to a point in the plane.

a � 2ba and b

b�2ba

a � ��2b�

FIGURE 9

a

b

FIGURE 10

a_2b

a-2b

a�a1, a2�.a

a

a � �a1, a2 �

�a1, a2 �

EXAMPLE 2

�a1, a2�

FIGURE 11 a=ka¡, a™l

(a¡, a™)

O

y

x

a


For instance, the vectors shown in Figure 12 are all equivalent to the vector OPl

whose terminal point is . What they have in common is that the terminal point is reached from the initial point by a displacement of three units to the rightand two upward. We can think of all these geometric vectors as representations of the algebraic vector . The particular representation OP

lfrom the origin to the point

is called the position vector of the point .

P�3, 2�� 3, 2 �

a � �3, 2 �PP�3, 2�

FIGURE 12Representations of the vector a=k3, 2l

(1, 3)

(4, 5)

x

y

0

P(3, 2)

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Figure 13 shows a general vector with position vector OPl

and another rep-resentation AB

lof , where the initial point is and the terminal point is .

Then we must have , and , so , and .Thus we have the following result.

Given the points and , the vector with representation ABl

is

Find the vector represented by the directed line segment with initial pointand terminal point .

SOLUTION By , the vector corresponding to ABl

is

The magnitude or length of the vector is the length of any of its representations andis denoted by the symbol or . By using the distance formula to compute the lengthof a segment , we obtain the following formula.

The length of the vector is

How do we add vectors algebraically? Figure 14 shows that if and, then the sum is , at least for the case where the

components are positive. In other words, to add algebraic vectors we add their components.Similarly, to subtract vectors we subtract components. From the similar triangles in Figure15 we see that the components of are and . So to multiply a vector by a scalar wemultiply each component by that scalar.

a � �a1, a2 �B�x2, y2�A�x1, y1�a

a2 � y2 � y1a1 � x2 � x1y1 � a2 � y2x1 � a1 � x2

aB�x2, y2�A�x1, y1�1

a � �x2 � x1, y2 � y1 �

EXAMPLE 3B��2, 1�A�2, �3�

1

a � ��2 � 2, 1 � ��3�� 4, 4 �

v� v �� v �

OP

a � �a1, a2 �

� a � � sa 21 � a 2

2

a � �a1, a2 �a � b � �a1 � b1, a2 � b2 �b � �b1, b2 �

ca2ca1ca

FIGURE 14

0

y

xb¡a¡

b¡

a™ a™

b™ba+b

a

(a¡+b¡, a™+b™)

FIGURE 15

ca™

ca¡

caa™

a¡

a

FIGURE 13a=ka¡, a™l

(a¡, a™)

O

y

x

P

Representations of

A(⁄, ›)

position vector of P

B(¤, fi)


If and , then

ca � �ca1, ca2 �

a � b � �a1 � b1, a2 � b2 �a � b � �a1 � b1, a2 � b2 �

b � �b1, b2 �a � �a1, a2 �

If and , find and the vectors , , ,and .

SOLUTION

3ba � ba � b� a �b � ��2, 1 �a � �4, 3 �EXAMPLE 42a � 5b

� a � � s42 � 32 � s25 � 5

� �4 � 2, 3 � 1 � � �2, 4 �a � b � �4, 3 � � ��2, 1 �

� �4 � ��2�, 3 � 1 � � �6, 2 �a � b � �4, 3 � � ��2, 1 �

3b � 3 ��2, 1 � � �3��2�, 3�1�� 6, 3 �

� �8, 6 � � ��10, 5 � � ��2, 11 �2a � 5b � 2 �4, 3 � � 5 ��2, 1 �

SECTION A.1 VECTORS IN TWO DIMENSIONS A5

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Properties of Vectors If , , and are vectors and and are scalars, then

1. 2.

3. 4.

5. 6.

7. 8.

These eight properties of vectors can be readily verified either geometrically or alge-braically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the Paral- lelogram Law) or as follows:

We can see why Property 2 (the associative law) is true by looking at Figure 16 andapplying the Triangle Law several times: The vector PQ

lis obtained either by first con-

structing a � b and then adding c or by adding a to the vector b � c.Two vectors play a special role. Let

These vectors and are called the standard basis vectors. They have length and pointin the directions of the positive - and -axes. (See Figure 17.)

If , then we can write

dccba

a � �b � c� � �a � b� � ca � b � b � a

a � ��a� � 0a � 0 � a

�c � d �a � ca � dac�a � b� � ca � cb

1a � a�cd �a � c�da�

a � b � �a1, a2 � � �b1, b2 � � �a1 � b1, a2 � b2 �

� �b1 � a1, b2 � a2 � � �b1, b2 � � �a1, a2 �

� b � a

j � �0, 1 �i � �1, 0 �

1jiyx

a � �a1, a2 �

� a1 �1, 0 � � a2 �0, 1 �a � �a1, a2 � � �a1, 0 � � �0, a2 �

a � a1 i � a2 j2

A6

FIGURE 16

b

c

a

(a+b)+c

P

Q

=a+(b+c)a+b

b+c

FIGURE 17Standard basis vectors

0

y

x

j

(1, 0)

i

(0, 1)


Thus any vector can be expressed in terms of and . For instance,

See Figure 18 for the geometric interpretation of Equation 2 and compare with Figure 17.

If and , express the vector in terms of and .

SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have

2a � 3b � 2�i � 2 j� � 3�4 i � 7j�

� 2 i � 4 j � 12 i � 21j � 14 i � 25 j

2a � 3bb � 4 i � 7ja � i � 2 jEXAMPLE 5ji

ji

�1, �2 � � i � 2 j

FIGURE 18

a=a¡i+a™ j

0

y

x

a

a¡i

a™ j

(a¡, a™)

APPENDIX A VECTORS

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A7

A unit vector is a vector whose length is 1. For instance, and are both unit vectors.In general, if , then the unit vector that has the same direction as is

In order to verify this, we let . Then and is a positive scalar, so hasthe same direction as . Also

Find the unit vector in the direction of the vector .

SOLUTION The given vector has length

so, by Equation 3, the unit vector with the same direction is

Applications

A force is represented by a vector because it has both a magnitude (measured in poundsor newtons) and a direction. If several forces are acting on an object, the resul tant forceexperienced by the object is the vector sum of these forces.

A 100-lb weight hangs from two wires as shown in Figure 19. Find the ten-sions (forces) and in both wires and their magnitudes.

SOLUTION We first express and in terms of their horizontal and vertical compo-nents. From Figure 20 we see that

The resultant of the tensions counterbalances the weight and so we must have

i ja � 0 a

u �1

� a � a �a

� a �c � 1�� a � u � ca c u

a

� u � � � ca � � � c � � a � �1

� a � � a � � 1

3 i � 4 j

� 3 i � 4 j � � s32 � ��4�2 � s25 � 5

15 �3 i � 4 j� � 3

5 i �45 j

T1 T2

T1 T2

T1 � �� T1 � cos 50� i � � T1 � sin 50� j

T2 � � T2 � cos 32� i � � T2 � sin 32� j

T1 � T2 w

T1 � T2 � �w � 100 j

3

EXAMPLE 6

EXAMPLE 7

4

5

Gibbs

Josiah Willard Gibbs (1839–1903), a professorof mathematical physics at Yale College, pub-lished the first book on vectors, Vector Analysis,in 1881. More complicated objects, calledquaternions, had earlier been invented byHamilton as mathematical tools for describingspace, but they weren’t easy for scientists touse. Quaternions have a scalar part and a vec-tor part. Gibb’s idea was to use the vector partseparately. Maxwell and Heaviside had similarideas, but Gibb’s approach has proved to be themost convenient way to study space.

FIGURE 20

50°

w

T¡

50° 32°

32°

T™

FIGURE 19

100

T¡

50° 32°

T™

48238_ch10_ptg01_hr_680-689_48238_ch10_ptg01_hr_680-689 1/24/11 5:20 PM 83

Vectors are useful in many aspects of physics and engineering. In Section A.3 we will see how they describe the velocity and acceleration of objects moving along a curve. Here we look at forces.

Thus

Equating components, we get

Solving the first of these equations for and substituting into the second, we get

(�� T1 � cos 50� � � T2 � cos 32�) i � (� T1 � sin 50� � � T2 � sin 32�) j � 100 j

�� T1 � cos 50� � � T2 � cos 32� � 0

� T1 � sin 50� � � T2 � sin 32� � 100

� T2 �

� T1 � sin 50� � � T1� cos 50�

cos 32�sin 32� � 100

SECTION A.1 VECTORS IN TWO DIMENSIONS

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A8

1. Are the following quantities vectors or scalars? Explain.(a) The cost of a theater ticket(b) The current in a river(c) The initial flight path from Houston to Dallas(d) The population of the world

2. What is the relationship between the point (4, 7) and the vector ? Illustrate with a sketch.

3. Name all the equal vectors in the parallelogram shown.

4. Write each combination of vectors as a single vector.

(a) ABl

BCl

(b) CDl

DBl

(c) DBl

ABl

(d) DCl

CAl

ABl

�4, 7 �

B

E

A

D C

��

��

6. Copy the vectors in the figure and use them to draw the following vectors.(a) (b)(c) (d)(e) (f )

7. In the figure, the tip of and the tail of are both the midpointof . Express and in terms of and .

a � ba � b�3b1

2 a2b � aa � 2b

b a

dcbadcQR

ba c

d

P

Q

R

A.1 Exercises

So the magnitudes of the tensions are

and

Substituting these values in and , we obtain the tension vectors

� T1 � �100

sin 50� � tan 32� cos 50� 85.64 lb

� T2 � � � T1 � cos 50�

cos 32� 64.91 lb

T1 �55.05 i � 65.60 j T2 55.05 i � 34.40 j

4 5


8. If the vectors in the figure satisfy and, what is ?

�u � � �v � � 1u � v � w � 0 �w �

u

v

w

APPENDIX A VECTORS

A

DC

B

5. Copy the vectors in the figure and use them to draw the following vectors.(a) (b)(c) (d)(e) (f )

u � wu � vu � vv � wu � w � vv � u � w

wvu

9–12 Find a vector with representation given by the directed linesegment AB

l. Draw AB

land the equivalent representation starting at

the origin.

9. , 10. ,

11. , 12. ,

a

A��1, 1� B�3, 2� A��4, �1� B�1, 2�

A��1, 3� B�2, 2� A�2, 1� B�0, 6�

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A9

13–16 Find the sum of the given vectors and illustrate geometrically.

13. , 14. ,

15. , 16. ,

17–24 Find , a � b, a � b, , and .

17. ,

18. ,

19. , 20. ,

21. , 22. ,

23. , 24. ,

25–27 Find a unit vector that has the same direction as the givenvector.

25. 26.

27.

28. Find a vector that has the same direction as but haslength 5.

29–30 What is the angle between the given vector and the positivedirection of the -axis?

29. 30.

31. If lies in the first quadrant and makes an angle with thepositive -axis and , find in component form.

32. If a child pulls a sled through the snow with a force of 50 Nexerted at an angle of above the horizontal, find thehorizontal and vertical components of the force.

33. A quarterback throws a football with angle of elevation andspeed . Find the horizontal and vertical components ofthe velocity vector.

��1, 5 ��3, �1 ��6, �2 ��1, 4 �

�5, 3 ��1, 2 ��3, �4 ��2, 3 �

3a � 4b2a� a �b � ��2, 8 �a � �5, �12 �

b � �4, 3 �a � ��1, 2 �

b � �1, 4 �a � �2, �3 � b � �6, 7 �a � �2, �1 �

b � i � ja � i � j b � 3i � 2 ja � 2 i � 3 j

b � 2 i � ja � i � j b � i � 2 ja � 6 i

�3, �5 ��1, 2 �

i � j

��2, 4 �

x

8 i � 6 ji � s3 j

��3vv� v � � 4x

38 �

40 �60 ft�s

the direction of the resul tant of the velocity vectors of the planeand the wind. The ground speed of the plane is the magnitudeof the resultant. Find the true course and the ground speed ofthe plane.

37. A woman walks due west on the deck of a ship at 3 mi�h. Theship is moving north at a speed of 22 mi�h. Find the speed anddirection of the woman relative to the surface of the water.

38. Ropes 3 m and 5 m in length are fastened to a holiday decora-tion that is suspended over a town square. The decoration has amass of 5 kg. The ropes, fastened at different heights, makeangles of and with the horizontal. Find the tension ineach wire and the magnitude of each tension.

39. A clothesline is tied between two poles, 8 m apart. The line is quite taut and has negligible sag. When a wet shirt with amass of 0.8 kg is hung at the middle of the line, the mid point is pulled down 8 cm. Find the tension in each half of theclothesline.

40. The tension T at each end of the chain has magnitude 25 N(see the figure). What is the weight of the chain?

41. Find the unit vectors that are parallel to the tangent line to theparabola at the point .

42. (a) Find the unit vectors that are parallel to the tangent line tothe curve at the point .

(b) Find the unit vectors that are perpendicular to the tangentline.

40�52�

3 m 5 m

52°40°

37° 37°

�2, 4�y � x 2

��6, 1�y � 2 sin x


34–35 Find the magnitude of the resultant force and the angle itmakes with the positive -axis.

34. 35.

x

300 N

200 N

60°0

y

x

20 lb

16 lb

45°0

y

x30°

(c) Sketch the curve and the vectors in parts (a) and (b), all starting at .

43. If , , and are the vertices of a triangle, find

ABl

� BCl

� CAl

.

44. Let be the point on the line segment that is twice as far from as it is from . If OA

l, OB

l, and OC

l, show

that .

��6, 1�

CBA

ABCc �b �a �AB

c � 23 a �

13 b

y � 2 sin x

SECTION A.1 VECTORS IN TWO DIMENSIONS

36. Velocities have both direction and magnitude and thus are vectors. The magnitude of a velocity vector is called speed.Suppose that a wind is blowing from the direction N W at aspeed of 50 km�h. (This means that the direction from whichthe wind blows is west of the northerly direction.) A pilot is

45�

45�

45. (a) Draw the vectors , , and

(b) Show, by means of a sketch, that there are scalars andsuch that .

(c) Use the sketch to estimate the values of and .(d) Find the exact values of and .

b � �2, �1 �a � �3, 2 �c � �7, 1 � .

tsc � sa � tb

tstssteering a plane in the direction N E at an airspeed (speed in60�

still air) of 250 km�h. The true course, or track, of the plane is

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A10

46. Suppose that and are nonzero vectors that are not paralleland is any vector in the plane determined by and . Give a geometric argument to show that can be written as

babac

c

49. Figure 16 gives a geometric demonstration of Property 2 of vectors. Use components to give an algebraic proof of this fact.

50. Prove Property 5 of vectors algebraically. Then use similar triangles to give a geometric proof.

51. Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

In general, a function is a rule that assigns to each element in the domain an element in therange. A vector-valued function, or vector function, is simply a function whose domainis a set of real numbers and whose range is a set of vectors. This means that for every num-ber in the domain of a vector function there is a unique vector denoted by . If and

are the components of the vector , then and are real-valued functions called thecomponent functions of and we can write

We use the letter to denote the independent variable because it represents time in most appli-cations of vector functions.

If

then the component functions are

By our usual convention, the domain of consists of all values of for which the expres-sion for is defined. The expressions and are defined whenand . Therefore the domain of is the interval .

The limit of a vector function is defined by taking the limits of its component functionsas follows.

If , then

provided the limits of the component functions exist.

Equivalently, we could have used an definition. Limits of vector functions obey thesame rules as limits of real-valued functions.

t r r�t� f �t�t�t� r�t� f t

r

r�t� � � f �t�, t�t�� f �t� i � t�t� j

t

r�t� � �ln�3 � t�, st�

f �t� � ln�3 � t� t�t� � st

r tr�t� ln�3 � t� st 3 � t � 0

t � 0 r �0, 3�

r

r�t� � � f �t�, t�t��

limt l a

r�t� � � limt l a

f �t�, limt l a

t�t��

EXAMPLE 1

1

�-�

A.2 Vector Functions and Their Derivatives

If , this definition is equivalentto saying that the length and direction of thevector approach the length and direction ofthe vector .L

r�t�

lim t la r�t� � L

48238_ch10_ptg01_hr_680-689_48238_ch10_ptg01_hr_680-689 2/7/11 10:38 AM Page 686

for suitable scalars and Then give an argu-ment using components.

47. If and , describe the set of all pointssuch that .

c � sa � tb s t.

r � �x, y � r0 � �x0, y0 ��x, y� � r � r0 � � 1

APPENDIX A VECTORS

48. If , , and , describe the set of all points such that ,where .

r � �x, y � r1 � �x1, y1 � r2 � �x2, y2 ��x, y� � r � r1 � � � r � r2 � � k

k � � r1 � r2 �

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SECTION VECTOR FUNCTIONS AND THEIR DERIVATIVES A11

FIGURE 1 C is traced out by the tip of a movingposition vector r(t).

0

y

x

C

P{f(t), g(t)}

r(t)=kf(t), g(t)l

FIGURE 2

0

y

x

1

3

r(t)

0

y

x

2

3r(π) r(0)

FIGURE 3

Find , where .

SOLUTION According to Definition 1, the limit of r is the vector whose components arethe limits of the component functions of r:

(by Equation 2.4.2)

A vector function is continuous at a if

In view of Definition 1, we see that is continuous at if and only if its component func-tions and are continuous at .

There is a close connection between continuous vector functions and parametric curves.Suppose that and are continuous real-valued functions on an interval . Then the setof all points , where

and varies throughout the interval , is a parametric curve. We can think of as beingtraced out by a moving particle whose position at time is . If we now considerthe vector function , then is the position vector of the point

on . Thus any continuous vector function defines a curve that is traced outby the tip of the moving vector , as shown in Figure 1.

The curve defined by the vector function

is the

A.2

curve described by the parametric equations

Here we simply take a different point of view: We regard by the tip of the moving vector . (See Figure 2.)

Sketch the curve with vector equation , .

SOLUTION The corresponding parametric equations are

We eliminate the parameter by observing that

So the curve is the ellipse

shown in Figure 3. As increases from to , the position vector rotates oncecounterclockwise around the origin.

limt l 0

r�t� � limt l 0

�1 � t 3 �� i � �limt l 0

sin t

t j

� i � j

r

limt l a

r�t� � r�a�

r af t a

f t I C�x, y�

x � f �t� y � t�t�

t I Ct � f �t�, t�t��

r�t� � � f �t�, t�t�� r�t�P� f �t�, t�t�� C r C

r�t�

r�t� � � t 2 � 2t, t � 1 � � �t 2 � 2t�i � �t � 1�j

x � t 2 � 2t y � t � 1

r�t�

r�t� � �3 cos t, 2 sin t � 0 t 2�

x � 3 cos t y � 2 sin t

� x

3�2

� � y

2�2

� cos2t � sin2t � 1

EXAMPLE 3

x 2

9�

y2

4� 1

EXAMPLE 4

r�t�2�0t

r�t� � �1 � t 3 � i �sin t

tjlim

t l 0r�t�EXAMPLE 2


the parabola as being traced out

Not For Sale

A12

Vector Equation of a LineA line is determined when a point on the line and the direction of the line are given. Thedirection of a line in the plane is determined by specifying a slope or by specifying a vec-tor that is parallel to the line. The latter case leads to the vector equation of the line. Letbe a vector parallel to . Let be an arbitrary point on and let and be the position

vectors of and (so OPA and OPA in Figure 4). If is the vector with repre-

sentation P PA, as in Figure 4, then the Triangle Law for vector addition gives .But, since and are parallel vectors, there is a scalar such that . Thus

which is a vector equation of . Each value of the parameter gives the position vectorof a point on . In other words, as varies, the line is traced out by the tip of the vec-tor . As Figure 5 indicates, positive values of correspond to points on that lie on one sideof , whereas negative values of correspond to points that lie on the other side of

If the vector that gives the direction of the line is written in component form as, then we have . We can also write and ,

so the vector equation becomes

Two vectors are equal if and only if corresponding components are equal. Therefore wehave the scalar equations:

where . These equations are parametric equations of the line through the pointand parallel to the vector . Each value of the parameter gives a point

on . Note that if is parallel to the line, then any nonzero multiple of is also par-allel to the line and can be used to obtain a vector equation or parametric equations of theline.

Find vector and parametric equations of the line that passes through thepoints and .

SOLUTION We are not explicitly given a vector parallel to the line but observe that the

vector with representation is parallel to the line and

Any nonzero multiple of can also be used. We can use any point on the line, for example , for the point The vector equation of the line becomes

or, equivalently,

Parametric equations of this line are

P0

vL P L r0 r

P0 P r0 � 0 r � a

0 r � r0 � aa v t a � tv

r � r0 � tv

L t rL t

r t LP0 t P0.

v Lv � �a, b � tv � � ta, tb � r � �x, y � r0 � �x0, y0 �

�x, y � � �x0 � ta, y0 � tb �

x � x0 � at y � y0 � bt

t � � LP0�x0, y0� v � �a, b � t�x, y� L v v

A�3, �2� B�5, 7�

v ABl

v � �5, 7 � � �3, �2 � � �2, 9 �

v�3, �2� P0.

r�t� � r0 � tv � �3, �2 � � t �2, 9 �

2

3

EXAMPLE 5

r�t� � �3 � 2t�i � ��2 � 9t�j

y � �2 � 9tx � 3 � 2t

2

xO

y

a

vr

r¸L

P¸(x¸, y¸)

P(x, y )

FIGURE 4

x

y

Lt=0 t>0

t<0

r¸

FIGURE 5

O


APPENDIX A VECTORS

Not For Sale

NOTE If we restrict the parameter as in Example 5 so that , then the para-metric equations , represent the line segment that starts atand ends at . Different parametric representations of the same line can be obtained by choosing to be or any other point on the line. Likewise, we can use any mul-tiple of , such as , , or .

Derivatives of Vector Functions

The derivative of a vector function is defined in much the same way as for real-valued functions:

if this limit exists. The geometric significance of this definition is shown in Figure 6. If thepoints and have position vectors and , then PQ

lrepresents the vector

, which can therefore be regarded as a secant vector. If , the scalarmultiple has the same direction as . As , itappears that this vector approaches a vector that lies on the tangent line. For this reason, thevector is called the tangent vector to the curve defined by at the point , providedthat exists and . The tangent line to at is defined to be the line through parallel to the tangent vector . We will also have occasion to consider the unit tangentvector, which is

The following theorem gives us a convenient method for computing the derivative of avector function ; just differentiate each component of .

Theorem If , where and are differen-tiable functions, then

PROOF

x � 3 � 2t y � �2 � 9t �3, �2��5, 7�P0 �5, 7�

v �4, 18 � �1, 4.5 � ��6, �27 �

r� r

drdt

� r��t� � limh l 0

r�t � h� � r�t�h

P Q r�t� r�t � h�r�t � h� � r�t� h � 0

�1�h��r�t � h� � r�t�� r�t � h� � r�t� h l 0

r��t� r Pr��t� r��t� � 0 C P P

r��t�

T�t� �r��t�

� r��t� �

r r

r�t� � � f �t�, t�t�� f �t� i � t�t� j f t

r��t� � � f ��t�, t��t�� f ��t� i � t��t� j

r��t� � limh l 0

1

h�r�t � h� � r�t�

� limh l 0

1

h�� f �t � h�, t�t � h�� f �t�, t�t��

� limh l 0

f �t � h� � f �t�h

, t�t � h� � t�t�

h �� lim

h l 0

f �t � h� � f �t�h

, limh l 0

t�t � h� � t�t�h �

� � f ��t�, t��t��

0 � t � 1t

4

5

FIGURE 6

(b) The tangent vector

(a) The secant vector

0

z

x

P

C

Q

r(t+h)-r(t)

r(t)

r(t+h)

r(t+h)-r(t)

h

0

z

C

PQ

r(t+h)

r(t)

rª(t)

x


SECTION VECTOR FUNCTIONS AND THEIR DERIVATIVESA.2 A13

Not For Sale

(a) Find the derivative of .(b) Find the unit tangent vector at the point where .

SOLUTION(a) According to Theorem 5, we differentiate each component of r:

(b) Since and , the unit tangent vector at the point is

For the curve , find and sketch the position vec-tor and the tangent vector .

SOLUTION We have

Elimination of the parameter from the equations , gives ,. In Figure 7 we draw the position vector starting at the origin and the

tangent vector starting at the corresponding point .

Integrals of Vector Functions

The definite integral of a continuous vector function can be defined in much the sameway as for real-valued functions except that the integral is a vector. But then we can expressthe integral of in terms of the integrals of its component functions and as follows.

and so

This means that we can evaluate an integral of a vector function by integrating each com-ponent function.

We can extend the Fundamental Theorem of Calculus to continuous vector functions asfollows:

where is an antiderivative of , that is, . We use the notation for indefi-nite integrals (antiderivatives).

r�t� � �1 � t 3 � i � te�t jt � 0

r��t� � 3t 2 i � �1 � t�e�t j

r�0� � i r��0� � j �1, 0�

T�0� �r��0�

� r��0� � �j1

� j

r�t� � st i � �2 � t� j r��t�r�1� r��1�

r��t� �1

2sti � j and r��1� �

1

2 i � j

x � st y � 2 � t y � 2 � x 2

x � 0 r�1� � i � jr��1� �1, 1�

r�t�

r f t

yb

ar�t� dt � lim

n l ��n

i�1r�t*i � �t

� limn l �

��n

i�1f �t*i � �t� i � ��

n

i�1t�t*i � �t� j

yb

ar�t� dt � �y

b

af �t� dt� i � �y

b

at�t� dt� j

yb

ar�t� dt � R�t�]b

a � R�b� � R�a�

R r R��t� � r�t� x r�t� dt

EXAMPLE 6

EXAMPLE 7r(1) rª(1)

(1, 1)

FIGURE 7

0

y

2

x1

Notice from Figure 7 that the tangent vectorpoints in the direction of increasing . (SeeExercise 47.)

t


APPENDIX A VECTORSA14

Not For Sale

A15

1–2 Find the domain of the vector function.

1. 2.

3–6 Find the limit.

3. 4.

5. 6.

7–12 Sketch the curve with the given vector equation. Indicate withan arrow the direction in which increases.

7. 8.

9. 10.

11. 12.

13–15 Find (a) a vector equation and (b) parametric equations forthe line passing through the pair of points.

13. 14.

15.

16–18 Find (a) a vector equation, (b) parametric equations, and (c) a Cartesian equation for the line that passes through the pointand is parallel to the vector .

16. , 17. ,

18. ,

19. An object is moving in the -plane and its position after seconds is .

(a) Find the position of the object at time .(b) At what time is the object at the point ?(c) Does the object pass through the point ?(d) Find an equation in and whose graph is the path of

the object.

r�t� � st � 1, s5 � t � r�t� �t � 2

t � 2 i � ln�9 � t2� j

limt l 0�

sin t, t ln t � limt l 0

� e t � 1

t,

s1 � t � 1

t limt l 1

�st � 3 i �t � 1

t 2 � 1 j� lim

t l �arctan t, e�2t�

t

r�t� � t, 2t � r�t� � t 2, t �

r�t� � sin t i � cos t j r�t� � i � cos t j

r�t� � t 3 � 1, 2t � r�t� � t 2, t 3 �

�1, 3� and ��2, 7� ��3, 4� and �2, 8�

�4, �1� and ��2, 5�

Pa

P�1, 3� a � 1, �2 � P��4, 5� a � �2, 6 �

P�2, 5� a � 3, 0 �

xyt r�t� � �t 2 � 2t� i � �t � 4� j

t � 2�15, 7��20, 9�

yx

20. An object is moving in the -plane and its position after seconds is .

(a) Find the position of the object at time .(b) At what time is the object at the point ?(c) Does the object pass through the point ?(d) Find an equation in and whose graph is the path of

the object.

21. The figure shows a curve given by a vector function .(a) Draw the vectors and .(b) Draw the vectors

(c) Write expressions for and the unit tangent vector T(4).(d) Draw the vector T(4).

22. (a) Make a large sketch of the curve described by the vectorfunction , , and draw the vectorsr(1), r(1.1), and r(1.1) � r(1).

(b) Draw the vector starting at (1, 1) and compare it withthe vector

Explain why these vectors are so close to each other inlength and direction.

23–26 Find the domain and derivative of the vector function.

23. 24.

25. 26.

xyt r�t� � �t � 3� i � �t 2 � 2t� j

t � 5�1, 8��3, 20�

x y

C r�t�r�4.5� � r�4� r�4.2� � r�4�

r�4.5� � r�4�0.5

andr�4.2� � r�4�

0.2

r��4�

x0 1

1

yRC

Q

P

r(4.5)

r(4.2)

r(4)

r�t� � t 2, t � 0 � t � 2

r��1�

r�1.1� � r�1�0.1

r�t� � t 2, t � r�t� � t sin t i � t cos t j

r�t� � t 2 � 4, st � 4 � r�t� � � t 2

t � 1,

t

t 2 � 1

A.2 Exercises

If , then

where is a vector constant of integration, and

r�t� � 2 cos t i � sin t j

y r�t� dt � �y 2 cos t dt� i � �y sin t dt� j � 2 sin t i � cos t j � C

C

y�2

0r�t� dt � [2 sin t i � cos t j]0

�2� 2 i � j

EXAMPLE 8


SECTION VECTOR FUNCTIONS AND THEIR DERIVATIVESA.2

Not For Sale

A16

27–30 Find the tangent vector of unit length at the point with thegiven value of the parameter .

27. ,

28. ,

29. ,

30. ,

31–36(a) Sketch the plane curve with the given vector equation.(b) Find .(c) Sketch the position vector and the tangent vector for

the given value of .

31. ,

32. ,

33. ,

34. ,

35. ,

36. ,

37. Let .

(a) Find a vector tangent to the curve given by at the point where .

(b) Find parametric equations for the tangent line to the curveat the point where .

t

r�t� � 2t, 3t 2 � t � 1

r�t� � 2 sin t i � 3 cos t j t � �6

r�t� � �1 � t 3� i � t 2 j t � 1

r�t� � t cos t, t sin t � t � �4

r��t�r�t� r��t�

t

r�t� � e t i � e �t j t � 0

r�t� � e 2t i � e t j t � 0

r�t� � �t 2 � t� i �2

tj

r�t�t � 2

r�t� � t � 2, t 2 � 1 � t � �1

r�t� � t 2, t 3 � t � 1

r�t� � sin t i � 2 cos t j t � �4

t � 2

r�t� � �1 � cos t� i � �2 � sin t� j t � �6

38. Let .(a) Find a vector tangent to the curve given by at the

point .(b) Find parametric equations for the tangent line to the curve

at the point .

39–42 Evaluate the integral.

39. 40.

41. 42.

43. Find if and .

44. Find if and .

45–46 If is a real-valued function and is a vector-valued func-tion, where and are both differentiable, prove the differentia-tion formula.

45.

46.

47. Show that the tangent vector to a curve defined by a vectorfunction points in the direction of increasing . [Hint: Referto Figure 6 and consider the cases and separately.]

r�t� th 0 h � 0

��2, 4�

y1

0�16t3 i � 9t2 j� dt y

1

0� 4

1 � t 2 i �2t

1 � t 2 j� dt

y �e t i � ln t j� dt y �cos t i � sin t j� dt

r�t� r��t� � t 2 i � 4t 3 j r�0� � j

r�t� r��t� � sin t i � cos t j r�0� � i � j

f uf u

d

dt� f �t�u�t�� f ��t�u�t� � f �t�u��t�

d

dt�u� f �t�� f ��t�u�� f �t��

r�t� � �t 2 � t � 2� i � 4t 2 jr�t�

��2, 4�

A.3 Curvilinear Motion: Velocity and Acceleration

FIGURE 1

r(t+h)-r(t)

h

xO

y

C

PQ

rª(t)

r(t+h)r(t)

In this section we show how the ideas of derivatives and tangent vectors can be used inphys ics to study the motion of an object, including its velocity and acceleration, along acurve.

Suppose a particle moves so that its position vector at time is . Notice from Figure 1that, for small values of , the vector

approximates the direction of the particle moving along the curve . Its magnitude mea-sures the size of the displacement vector per unit time. The vector gives the averagevelocity over a time interval of length and its limit is the velocity vector at time :

Thus the velocity vector is also the tangent vector and points in the direction of the tangentline.

t r�t�h

r�t � h� � r�t�h

r�t�

h v�t� t

v�t� � limh l 0

r�t � h� � r�t�h

� r��t�

1

2

1


APPENDIX A VECTORS

Not For Sale

SECTION CURVILINEAR MOTION: VELOCITY AND ACCELERATION A17

The speed of the particle at time is the magnitude of the velocity vector, that is, .As in the case of one-dimensional motion, the acceleration of the particle is defined as thederivative of the velocity:

The position vector of an object moving in a plane is given byFind its velocity, speed, and acceleration when and illustrate

geometrically.

SOLUTION The velocity and acceleration at time are

and the speed is

When , we have

These velocity and acceleration vectors are shown in Figure 2.

The vector integrals that were introduced in Section 10.4 can be used to find position vec-tors when velocity or acceleration vectors are known, as in the following example.

A moving particle starts at an initial position with initialvelocity . Its acceleration is . Find its velocity and positionat time .

SOLUTION Since , we have

To determine the value of the constant vector , we use the fact that . Thepreceding equation gives , so and

Since , we have

Putting , we find that , so

t � v�t� �

a�t� � v��t� � r��t�

r�t� � t 3 i � t 2 j. t � 1

t

v�t� � r��t� � 3t 2 i � 2t j

a�t� � r��t� � 6t i � 2 j

� v�t� � � s�3t 2 �2 � �2t�2 � s9t 4 � 4t 2

t � 1

v�1� � 3 i � 2 j a�1� � 6 i � 2 j � v�1� � � s13

r�0� � 1, 0 �v�0� � i � j a�t� � 4t i � 6t j

t

a�t� � v��t�

v�t� � y a�t� dt � y �4t i � 6t j� dt

� 2t 2 i � 3t 2 j � C

C v�0� � i � jv�0� � C C � i � j

v�t� � 2t 2 i � 3t 2 j � i � j

� �2t 2 � 1� i � �3t 2 � 1� j

v�t� � r��t�

r�t� � y v�t� dt

� y ��2t 2 � 1� i � �3t 2 � 1� j� dt

� ( 23 t 3 � t) i � �t 3 � t� j � D

t � 0 D � r�0� � i

r�t� � ( 23 t 3 � t � 1) i � �t 3 � t� j

EXAMPLE 1

EXAMPLE 2

FIGURE 2

0

y

x

(1, 1)a(1)

v(1)

A.3


Not For Sale

A18

In general, vector integrals allow us to recover velocity when acceleration is known andposition when velocity is known:

If the force that acts on a particle is known, then the acceleration can be found from New-ton’s Second Law of Motion. The vector version of this law states that if, at any time , aforce acts on an object of mass producing an acceleration , then

An object with mass that moves in a circular path with constant angularspeed has position vector . Find the force acting on theobject and show that it is directed toward the origin.

SOLUTION

Therefore Newton’s Second Law gives the force as

Notice that . This shows that the force acts in the direction opposite tothe radius vector and therefore points toward the origin (see Figure 3). Such a forceis called a centripetal (center-seeking) force.

A projectile is fired with angle of elevation and initial velocity . (SeeFigure 4.) Assuming that air resistance is negligible and the only external force is due togravity, find the position function of the projectile. What value of maximizes therange (the horizontal distance traveled)?

SOLUTION We set up the axes so that the projectile starts at the origin. Since the forcedue to gravity acts downward, we have

where m�s . Thus

Since , we have

where . Therefore

Integrating again, we obtain

v�t� � v�t0� � yt

t0

a�u� du r�t� � r�t0� � yt

t0

v�u� du

tF�t� m a�t�

F�t� � ma�t�

m r�t� � a cos t i � a sin t j

v�t� � r��t� � �a sin t i � a cos t j

a�t� � v��t� � �a 2 cos t i � a 2 sin t j

F�t� � ma�t� � �m 2�a cos t i � a sin t j�

F�t� � �m 2r�t�r�t�

� v0

r�t� �

F � ma � �mt j

t � � a � � 9.8 2

a � �t j

v��t� � a

v�t� � �tt j � C

C � v�0� � v0

r��t� � v�t� � �tt j � v0

r�t� � �12 tt 2 j � t v0 � D

EXAMPLE 3

EXAMPLE 4

The angular speed of the object moving withposition is , where is the angleshown in Figure 3.

� � d��dtP

FIGURE 3

P

¨

y

x0

FIGURE 4

0

y

x

a

d

v¸


APPENDIX A VECTORS

Not For Sale

SECTION CURVILINEAR MOTION: VELOCITY AND ACCELERATION A19

But , so the position vector of the projectile is given by

If we write (the initial speed of the projectile), then

and Equation 3 becomes

The parametric equations of the trajectory are therefore

The horizontal distance is the value of when . Setting , we obtainor . The latter value of then gives

Clearly, has its maximum value when , that is, .

A projectile is fired with muzzle speed and angle of elevationfrom a position 10 m above ground level. Where does the projectile hit the ground, andwith what speed?

SOLUTION If we place the origin at ground level, then the initial position of the projectileis (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for .With , , and , we have

Impact occurs when , that is, . Solving this quadraticequation (and using only the positive value of ), we get

Then so the projectile hits the ground about 2306 m away.The velocity of the projectile is

So its speed at impact is

D � r�0� � 0

r�t� � �12 tt 2 j � t v0

� v0 � � v0

v0 � v0 cos � i � v0 sin � j

r�t� � �v0 cos ��t i � [�v0 sin ��t �12 tt 2 ] j

x � �v0 cos ��t y � �v0 sin ��t �12 tt 2

d x y � 0 y � 0 t � 0t � �2v0 sin ��t t

d � x � �v0 cos ��2v0 sin �

t�

v20�2 sin � cos ��

t�

v20 sin 2�

t

d sin 2� � 1 � � �4

150 m�s 45�

yv0 � 150 m�s � � 45� t � 9.8 m�s2

x � 150 cos��4�t � 75s2 t

y � 10 � 150 sin ��4� t �12 �9.8�t 2 � 10 � 75s2 t � 4.9t 2

y � 0 4.9t 2 � 75s2 t � 10 � 0t

t �75s2 � s11,250 � 196

9.8� 21.74

x � 75s2�21.74� � 2306,

v�t� � r��t� � 75s2 i � (75s2 � 9.8t) j

� v�21.74� � � s(75s2)2� (75s2 � 9.8 � 21.74)2 � 151 m�s

3

4

EXAMPLE 5

If you eliminate from Equations 4, you will seethat is a quadratic function of . So the pathof the projectile is part of a parabola.

xyt


A.3

Not For Sale

A20

1. The figure shows the path of a particle that moves withposition vector at time .(a) Draw a vector that represents the average velocity of the

particle over the time interval .(b) Draw a vector that represents the average velocity over the

time interval .(c) Write an expression for the velocity vector v(2).(d) Draw an approximation to the vector v(2) and estimate the

speed of the particle at .

2–4 The vector function represents the position of a particle attime . Find the velocity, speed, and acceleration at the given valueof .

2. ,

3. ,

4. ,

5–8 Find the velocity, acceleration, and speed of a particle with thegiven position function. Sketch the path of the particle and drawthe velocity and acceleration vectors for the specified value of .

5. ,

6. ,

7. ,

8. ,

9. If a ball is thrown into the air with a velocity of ,its position after seconds is given by

(a) Find the velocity of the ball when .(b) What is the speed of the ball when ?

r�t� t

2 � t � 2.4

1.5 � t � 2

t � 2

y

x0 21

2

1

r(2.4)

r(2)

r(1.5)

r�t�t

t

r�t� � 4 cos t, 3 sin t � t � �3

r�t� � �t 4 � 7t� i � �t 2 � 2t� j t � 1

r�t� � st 2 � 5 i � t j t � 2

t

r�t� � 2 � t, 4st � t � 1

10 i � 30 j ft�st

r�t� � 10t i � �30t � 16t 2� j

t � 1t � 1

r�t� � �12 t 2, t � t � 2

r�t� � 3 cos t i � 2 sin t j t � �3

r�t� � e t i � e 2 t j t � 0

10. If an arrow is shot upward on the moon with a velocity of, its position after seconds is given by

(a) Find the velocity of the arrow after 1 s.(b) Find the velocity of the arrow when .(c) When will the arrow hit the moon?(d) With what velocity will the arrow hit the moon?(e) With what speed will the arrow hit the moon?

11. A satellite completes one orbit of the earth along the equator at an altutude of 1000 km every 1 h 46 min. Find the velocity,speed, and acceleration of the satellite at any time. [Note: Theearth’s radius is approximately 6600 km.]

12. A satellite completes one orbit of the earth along the equator at an altutude of 2000 km every 2 h 8 min. Find the velocity,speed, and acceleration of the satellite at any time. [Note: Theearth’s radius is approximately 6600 km.]

13–14 Find the velocity and position vectors of a particle that has the given acceleration and the specified initial velocity andposition.

13. , ,

14. , ,

15–16(a) Find the position vector of a particle that has the given acceler-

ation and the specified initial velocity and position.

; (b) Use a calculator or computer to graph the path of the particle.

15. , ,

16. , ,

17. What force is required so that a particle of mass has the posi-tion function ?

18. A force with magnitude 20 N acts in the positive -direction onan object with mass 4 kg. The object starts at the origin withinitial velocity . Find its position function and itsspeed at time .

19. A projectile is fired with an initial speed of 200 m�s and angle of elevation . Find (a) the range of the projectile, (b) the maximum height reached, and (c) the speed at impact.

20. Rework Exercise 19 if the projectile is fired from a position100 m above the ground.

21. A ball is thrown at an angle of to the ground. If the balllands 90 m away, what was the initial speed of the ball?

15 i � 40 j m�s t

r�t� � 15t i � �40t � 0.83t 2� j

t � a

a�t� � 2 j v�0� � i � j r�0� � 0

a�t� � �10 j v�0� � i � j r�0� � 0

a�t� � 2 i � t j v�0� � i � j r�0� � i � j

a�t� � t 2 i � cos 2t j v�0� � j r�0� � i

mr�t� � t 3 i � t 2 j

y

v�0� � i � jt

60�

45�

A.3 Exercises

; Graphing calculator or computer required


APPENDIX A VECTORS

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22. A gun is fired with angle of elevation . What is the muzzle speed if the maximum height of the shell is 500 m?

23. A gun has muzzle speed . Find two angles of eleva-tion that can be used to hit a target 800 m away.

24. A batter hits a baseball 3 ft above the ground toward the cen-ter field fence, which is 10 ft high and 400 ft from homeplate. The ball leaves the bat with speed at an angle

above the horizontal. Is it a home run? (In other words,does the ball clear the fence?)

; 25. Water traveling along a straight portion of a river normallyflows fastest in the middle, and the speed slows to almostzero at the banks. Consider a long stretch of river flowingnorth, with parallel banks 40 m apart. If the maximum waterspeed is 3 , we can use a quadratic function as a basicmodel for the rate of water flow units from the west bank:

.

150 m�s

50�115 ft�s

m�sx

f �x� � 3400 x�40 � x�

30� (a) A boat proceeds at a constant speed of from a pointon the west bank while maintaining a heading perpen-

dicular to the bank. How far down the river on the oppo-site bank will the boat touch shore? Graph the path of theboat.

(b) Suppose we would like to pilot the boat to land at thepoint on the east bank directly opposite . If we main-tain a constant speed of and a constant heading,find the angle at which the boat should head. Then graphthe actual path the boat follows. Does the path seem realistic?

26. Another reasonable model for the water speed of the river inExercise 25 is a sine function: . If aboater would like to cross the river from to with constantheading and a constant speed of , determine the angleat which the boat should head.

AB5 m�s

f �x� � 3 sin�x�40�BA

5 m�s

A5 m�s


SECTION CURVILINEAR MOTION: VELOCITY AND ACCELERATIONA.3 A21

Not For Sale

A22

EXERCISES A.1 N PAGE A8

1. (a) Scalar (b) Vector (c) Vector (d) Scalar

3. ABl

� DCl

, DAl

� CBl

, DEl

� EBl

, EAl

� CEl

5. (a) (b)

(c) (d)

(e) (f)

7. ,

9. 11.

13. 15.

17.

19.

21.

23.

25. 27.

29. 31.33. , 35.37.39.41. 43.45. (a), (b) (d)

47. A circle with radius 1, centered at

w

u

u+wvu+v

u

_vu

u-v

w v

v+w

_wu

_v

u-w-vw

v

u

v+u+w

d � 12 b �

12 ac � 1

2 a �12 b

a � �3, �1 a � �4, 1

x0

y

A(_1, 3)

B(2, 2)

ax

y

A(_1, 1)

0

B(3, 2)

a

�5, �1 �5, 2

k5, _1l

k3, _4lk2, 3l

x0

y

k6, _2l

k5, 2l

k_1, 4l

13, �3, �4 , �7, �20 , �10, �24 , �7, �4 s13 , �3, 1 , �1, �7 , �4, �6 , �10, 7 s2 , 2 i, �2 j, 2 i � 2 j, 7 i � j

s2, 3 i, �i � 2 j, 2 i � 2 j, 11 i � j

�i � j��s2�1�s5, 2�s5 �2, 2s3 60�

�38.57 ft�s� 45.96 ft�s100s7 � 264.6 N, �139.1�

s493 � 22.2 mi�h, N8�WT1 � �196 i � 3.92 j, T2 � 196 i � 3.92 j

0��i � 4 j��s17s � 9

7 , t � 117

y

x0

a

b

c

sa

tb

�x0, y0 �

48238_Ans_ptg01_hr_A102-A111_48238_Ans_ptg01_hr_A102-A111 1/25/11 11:43 AM Page 102

ANSWERS TO VECTOR EXERCISES

EXERCISES N PAGE

1. [1, 5] 3. 5.7. 9.

11.

13. (a) (b) 15. (a) (b) 17. (a) (b) (c) 19. (a) (b) (c) No (d) 21. (a) (b), (d)

(c) ;

23.

25.

27. 29.

31. (a), (c) (b)

x

y

0

≈+¥=1

x

y

0 1

y=2x

1

x

y

0 2

2

x= Á+118

r�t� � �1, 3 � � t ��3, 4 �x � 1 � 3t, y � 3 � 4t

r�t� � �4, �1 � � t ��6, 6 �x � 4 � 6t, y � �1 � 6t

r�t� � ��4, 5 � � t ��2, 6 �y � �3x � 7x � �4 � 2t, y � 5 � 6t

x � y 2 � 6y � 8t � 3�8, 6�

y

x0 1

1

RC

Q

P

r(4.5)

r(4.2)

r(4)

r(4.5)-r(4)0.5

r(4.2)-r(4)0.2

T(4)

y

x0 1

1

RC

Q

P

r(4.5)

r(4.2)

r(4)

r(4.5)-r(4)

r(4.2)-r(4)

T�4� �r��4�

� r��4� �r��4� � limh l 0

r�4 � h� � r�4�h

2 i �12 j�0, 0 �

��, �� , r��t� � �2t, 1 �

�4, ��, r��t� � �2t, 1��2st � 4 � ��3�s13, 2�s13 ��1�s10, 3�s10 �

r��t� � �1, 2t �y

0 x

r(_1)rª(_1)

(_3, 2)

A.2 A15

A.4 Answers to ExercisesVector

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A

33. (a), (c) (b)

35. (a), (c) (b)

37. (a) (b)

39. 41.43.

EXERCISES N PAGE

1. (a), (b) (c)

(d)

3. , ,

5.

7.

9. (a) (b)

11.

y

0 xr ” ’

” , œ„2’œ„22

π4

rª ” ’π4

r��t� � cos t i � 2 sin t j

x

y

0

rª(0)

r(0)

(1, 1)

r��t� � 2e 2 t i � e t j

5 i �12 j x � 6 � 5t, y � 1 �

12 t, t � 0

4 i � 3 j e t i � �t ln t � t�j � C13 t 3 i � �t 4 � 1�j

limh l 0

r�2 � h� � r�2�h

y

x

1

2

r(2.4)

r(2)

r(1.5)

(a) (b)y

x210

1

2

r(2.4)

r(2)

r(1.5)

v(2)

11 i � 4 j s137 12 i � 2 j

v�t� � ��t, 1 �(_2, 2)

0

y

x

v(2)

a(2)

a�t� � ��1, 0 �� v�t� � � st 2 � 1

0

y

x

v ” ’π3

a ” ’π3

” , œ„3’32

(0, 2)

(3, 0)

v�t� � �3 sin t i � 2 cos t ja�t� � �3 cos t i � 2 sin t j

� v�t� � � s5 sin2 t � 4

s104 ft�s10 i � 2 j

v�t� ��27,030 sin 3.56t, 27,030 cos 3.56t� ,

� v � 27,030 km�h,a�t� ��96,131 cos 3.56t, �96,131 sin 3.56t �

48238_Ans_ptg01_hr_A102-A111_48238_Ans_ptg01_hr_A102-A111 2/8/11 3:27 PM Page 103

23

A.3 A20

ANSWERS TO VECTOR EXERCISES

13.

15. (a) (b)

17.19. (a) (b) (c) 21. 23. , 25. (a) 16 m (b) upstream

v�t� � i � �2t � 1�j, r�t� � t i � �t 2 � t�j

r�t� � �t 2 � t � 1� i � ( 16 t 3 � t � 1) j

8

�1

�1 20

(1, 1) t=0

F�t� � 6mt i � 2m j�3535 m �1531 m 200 m�s

30 m�s �10.2� �79.8��23.6�

40

_4

0

4

_12

0

12

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a2 appendix a vectors - cengage · pdf filea2 appendix a vectors the term vector is used by...

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